03/26/2014, 01:23 PM

Considering the thread

http://math.eretrandre.org/tetrationforu...hp?tid=844

It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.

The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.

Lef f(a)= b.

Then there must exist for most such a and b :

f(a2)=f(a3)=f(a)=b

where a2 and a3 are lineair indep. (real lineair indep of course )

then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.

( the superfunctions equations are in the directions a-a2 and a-a3 )

Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.

regards

tommy1729

http://math.eretrandre.org/tetrationforu...hp?tid=844

It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.

The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.

Lef f(a)= b.

Then there must exist for most such a and b :

f(a2)=f(a3)=f(a)=b

where a2 and a3 are lineair indep. (real lineair indep of course )

then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.

( the superfunctions equations are in the directions a-a2 and a-a3 )

Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.

regards

tommy1729